RBSE Solutions For Class 7 Maths Chapter 6 – The Triangle and its Properties

Exercise 6.1 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

Access Answers to NCERT Class 7 Maths Chapter 6 – The Triangle and its Properties Exercise 6.2

1. Find the value of the unknown exterior angle x in the following diagram:

(i)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50^o + 70^o

= x = 120^o

(ii)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 65^o + 45^o

= x = 110^o

(iii)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30^o + 40^o

= x = 70^o

(iv)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 60^o + 60^o

= x = 120^o

(v)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50^o + 50^o

= x = 100^o

(vi)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30^o + 60^o

= x = 90^o

2. Find the value of the unknown interior angle x in the following figures:

(i)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 50^o = 115^o

By transposing 50^o from LHS to RHS it becomes – 50^o

= x = 115^o – 50^o

= x = 65^o

(ii)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= 70^o + x = 100^o

By transposing 70^o from LHS to RHS it becomes – 70^o

= x = 100^o – 70^o

= x = 30^o

(iii)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^o.

= x + 90^o = 125^o

By transposing 90^o from LHS to RHS it becomes – 90^o

= x = 125^o – 90^o

= x = 35^o

(iv)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 60^o = 120^o

By transposing 60^o from LHS to RHS it becomes – 60^o

= x = 120^o – 60^o

= x = 60^o

(v)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^o.

= x + 30^o = 80^o

By transposing 30^o from LHS to RHS it becomes – 30^o

= x = 80^o – 30^o

= x = 50^o

(vi)

Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90^o.

= x + 35^o = 75^o

By transposing 35^o from LHS to RHS it becomes – 35^o

= x = 75^o – 35^o

= x = 40^o